Assume that the weight X in ounces of a “10- ounce” box of

Chapter 8, Problem 2E

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QUESTION:

Assume that the weight \(X\) in ounces of a "10ounce" box of cornflakes is \(N(\mu, 0.03)\). Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from this distribution.

(a) To test the hypothesis \(H_{0}: \mu \geq 10.35\) against the alternative hypothesis \(H_{1}: \mu<10.35\), what is the critical region of size \(\alpha=0.05\) specified by the likelihood ratio test criterion? HinT: Note that if \9\mu \geq 10.35\) and \(\bar{x}<10.35\), then \(\widehat{\mu}=10.35\).

(b) If a random sample of \(n=50\) boxes yielded a sample mean of \(bar{x}=10.31\), is \(H_{0}\) rejected? Hint: Find the critical value \(z_{\alpha}\) when \(H_{0}\) is true by taking \(\mu=10.35\), which is the extreme value in \(\mu \geq 10.35\).

(c) What is the \(p\)-value of this test?

Equation Transcription:



.

Text Transcription:

X

N(mu,0.03)

X_1,X_2,,X_n  

H_0:10.35

H_1:<10.35

alpha=0.05

mu > or = 10.35

Bar x <10.35

Wide hat mu=10.35  

n=50

Bar x=10.31

H_0

z_alpha

mu =10.35

mu > or = 10.35 .

p

Questions & Answers

QUESTION:

Assume that the weight \(X\) in ounces of a "10ounce" box of cornflakes is \(N(\mu, 0.03)\). Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from this distribution.

(a) To test the hypothesis \(H_{0}: \mu \geq 10.35\) against the alternative hypothesis \(H_{1}: \mu<10.35\), what is the critical region of size \(\alpha=0.05\) specified by the likelihood ratio test criterion? HinT: Note that if \9\mu \geq 10.35\) and \(\bar{x}<10.35\), then \(\widehat{\mu}=10.35\).

(b) If a random sample of \(n=50\) boxes yielded a sample mean of \(bar{x}=10.31\), is \(H_{0}\) rejected? Hint: Find the critical value \(z_{\alpha}\) when \(H_{0}\) is true by taking \(\mu=10.35\), which is the extreme value in \(\mu \geq 10.35\).

(c) What is the \(p\)-value of this test?

Equation Transcription:



.

Text Transcription:

X

N(mu,0.03)

X_1,X_2,,X_n  

H_0:10.35

H_1:<10.35

alpha=0.05

mu > or = 10.35

Bar x <10.35

Wide hat mu=10.35  

n=50

Bar x=10.31

H_0

z_alpha

mu =10.35

mu > or = 10.35 .

p

ANSWER:

Step 1 of 4

The electronic configuration of chlorine is  implying it has seven valence electrons and oxygen with its electronic configuration of  has six valence electrons

 

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